Mathematics - Continuity - Differentiability Question with Solution | TestHub

Given,

$f\left(x\right)=\left\{\begin{array}{cc}\left|4x^2-8x+5\right|,& \mathrm{if} 8x^2-6x+1\ge 0\\ \left[4x^2-8x+5\right],& \mathrm{if} 8x^2-6x+1<0\end{array}\right.$

$=\left\{\begin{array}{cc}4x^2-8x+5,& \mathrm{if} x\in \left[-\infty ,\frac{1}{4}\right]\cup \left[\frac{1}{2},\infty \right)\\ \left[4x^2-8x+5\right]& \mathrm{if} x\in \left(\frac{1}{4},\frac{1}{2}\right)\end{array}\right.$

$\Rightarrow f\left(x\right)=\left\{\begin{array}{cc}4x^2-8x+5& \mathrm{if} x\in \left(-\infty ,\frac{1}{4}\right]\cup \left[\frac{1}{2},\infty \right)\\ 3& x\in \left(\frac{1}{4},\frac{2-\sqrt{2}}{2}\right)\\ 2& x\in \left[\frac{2-\sqrt{2}}{2},\frac{1}{2}\right)\end{array}\begin{array}{}\\ \\ \end{array}\right.$

Now plotting the graph of $f\left(x\right)$ we get,

We can see $f\left(x\right)$ is not differentiable at $\frac{1}{4},x_1,\frac{1}{2}$ (where$x_1=\frac{2-\sqrt{2}}{2}$)